do exist five different number such as $a_1,a_2,a_3,a_4,a_5\in\mathbb{N}$ which sum of every three of them be divisible by the sum of two other?
I guess I have to prove
$i<j \implies a_i<a_j$
$a_4+a_5\le a_1+a_2+a_3$ is not possible.
but I don't know how to prove it
Assume w.n.l.g. that $a_1<a_2<a_3<a_4<a_5$. The two largest account for more than $2/5$ of the sum, so the three smallest account for less than $3/5$, and so $(a_1 + a_2 + a_3) / (a_4 + a_5) < 3/2$ and can't be any integer besides $1$. That is, $a_1+a_2+a_3 = a_4+a_5$. Now consider $(a_1+a_2+a_4)/(a_3+a_5)$. If this is $1$ as well, then $a_3=a_4$, which can't be true. So it's at least $2$: $a_1+a_2+a_4 \ge 2a_3 + 2a_5$. Now we have $$ -a_3+a_4 = a_1+a_2-a_5 > a_1+a_2-2a_5 \ge 2a_3 - a_4, $$ or $a_4 \ge (3/2)a_3$. But then $a_4+a_5 > 2a_4 \ge 3a_3 > a_1+a_2+a_3$, which contradicts the equality $a_4+a_5=a_1+a_2+a_3$ that we already proved.